Computing Cohomology via Sheaves for Projective Spaces I'm trying to prove that given spaces $\mathbb{RP}^n$, and abelian group $A$ and $\mathcal{A}$ the constant sheaf over $\mathbb{RP}^n$, that one can prove
$$
H^1\left (\mathbb{RP}^n, \mathcal{A}\right ) \;\; =\;\; \begin{cases}
A, & \text{if} \; n = 1 \\
A/2A, & \text{if} \; n\geq 2
\end{cases}
$$
Where $A/2A = \{a \in A \; | \; 2a =0\}$.  Here cohomology sets $H^1$ are defined as isomorphism classes of torsors over their base.  I'm reading Manifolds, Sheaves, and Cohomology by Wedhorn and the illustrative example was given for $\mathbb{S}^1$ where they showed $H^1\left (\mathbb{S}^1, \mathcal{A}\right ) = A$.  I followed their argument for $\mathbb{S}^1$ and was convinced that it applied to $\mathbb{RP}^1$, but I'm struggling to see if it's valid in the general case.  Here's my attempt:
If we can find an open cover $\mathcal{U}$ of simply connected neighborhoods then $H^1(\mathcal{U}, \mathcal{A}) \cong H^1\left (\mathbb{RP}^n, \mathcal{A}\right )$.  This is quite simple because $\mathbb{RP}^n$ is a manifold and has a covering of charts $\mathcal{U} = \{U_1, \ldots, U_{n+1}\}$ where
$$
U_i \;\; =\;\; \{[v_1,\ldots, v_{n+1}] \; : \; v_i \neq 0\}
$$
which are all diffeomorphic to $\mathbb{R}^n$.  We then have that $\mathcal{A}(U_i) = A$ since $U_i$ is connected and locally constant functions are simply just constant functions.  Now, shouldn't it be that $\mathcal{A}(U_i \cap U_j) \cong A\times A$ since the nontrivial intersection cleaves the neighborhoods such that $U_i\cap U_j \approx \mathbb{R}^n \sqcup \mathbb{R}^n$?  If this is correct, then the restriction map $\mathcal{A}(U_i) \to \mathcal{A}(U_i \cap U_j)$ should be $a \to (a,a)$.  We then have that $(a,b)$ and $(a',b')$ should be cohomologous if there are $(c,c)$ and $(d,d)$ such that
$$
(a,b) \;\; =\;\; (c,c) + (a',b') + (d,d) \;\; =\;\; (a' + c + d, b' + c + d) \;\; \sim \;\; (a' + e, b' + e)
$$
which implies that $H^1\left (\mathcal{U}, \mathcal{A}\right ) \cong A$?  This is clearly wrong, so I have a gap in my conceptual understanding somewhere.  Please help me in finding my error.  If at all possible (likely a separate question) I would appreciate some guiding principles to tackle proving $H^1\left (\mathbb{CP}^n, \mathcal{A}\right ) = 0$.
Edit #1
Ok, I realized that there was a big problem with my setup from before because I misconstruing the definition for cocycle with the 1-dimensional example.  What we need in the case of $\mathbb{RP}^n$ is a sequence of elements $(a_{ij}, b_{ij}) \in \mathcal{A}(U_i \cap U_j)$ such that we have
$$
(a_{ij}, b_{ij}) + (a_{jk}, b_{jk}) \;\; =\;\; (a_{ik}, b_{ik}).
$$
Two cocycles $\left \{ \left (a_{ij}, b_{ij} \right )\right \}$ and $\left \{\left ( a_{ij}', b_{ij}' \right ) \right \}$ are cohomologous  if there are elements $c_1, \ldots, c_{n+1} \in A$ such that
$$
(c_i, c_i) + (a_{ij}, b_{ij}) \;\; =\;\; \left (a_{ij}', b_{ij}' \right ) + (c_j, c_j)
$$
but a little algebraic manipulation shows that we simply require that for all $i,j$ we have $a_{ij} - b_{ij} = a_{ij}' - b_{ij}'$ which implies that any cocycles are cohomologous if they are $A$-translates of one another.  The question then becomes, what are all of these isomorphism classes?  What are the different types of cycles that can exist that are not $A$-translates of each other?
Edit #2
Given Eric Wofsey's comment below, I now recognize that I need to be specific about the representation of $\mathbb{RP}^n$ being employed.  I'm going to use the shorthand $U_{ij}:= U_i\cap U_j$ and $U_{ijk} := U_i \cap U_j \cap U_k$  I'm therefore going to refer to elements $f\in \mathcal{A}(U_{ij})$ as the pairs $\left (f_{ij}^{++}, f_{ij}^{+-}\right )$ where we recognize that $U_i\cap U_j$ is equal to a disjoint union $W_{ij}^{++} \sqcup W_{ij}^{+-}$ where $++$ indicates that the $i$th and $j$th coordinates agree in sign and $+-$ indicates they differ in sign.
When we move onto $U_{ijk}$ we recognize that there is one more cleaving and $U_{ijk} = W_{ijk}^{+++} \sqcup W_{ijk}^{-++} \sqcup W_{ijk}^{+-+}\sqcup W_{ijk}^{++-}$ and that $\mathcal{A}(U_{ijk}) \cong A\times A\times A\times A$.  We therefore find that when we restrict these maps $U_{ij} \to U_{ijk}$ we get the splittings
\begin{eqnarray*}
\left (f_{ij}^{++}, f_{ij}^{+-}\right ) & \to & \left (f_{ij}^{++}, f_{ij}^{+-}, f_{ij}^{+-}, f_{ij}^{++}\right ) \\
\left (f_{jk}^{++}, f_{jk}^{+-}\right ) & \to & \left (f_{jk}^{++}, f_{jk}^{++}, f_{jk}^{+-}, f_{jk}^{+-}\right ) \\
\left (f_{ik}^{++}, f_{ik}^{+-}\right ) & \to & \left (f_{ik}^{++}, f_{ik}^{+-}, f_{ik}^{++}, f_{ik}^{+-}\right ). 
\end{eqnarray*}
From here we find the (hopefully) correct cocycle condition:
\begin{eqnarray*}
f_{ij}^{++} + f_{jk}^{++} & = & f_{ik}^{++} \hspace{1pc} \text{on} \;\; W_{ijk}^{+++} \\
f_{ij}^{+-} + f_{jk}^{++} & = & f_{ik}^{+-} \hspace{1pc} \text{on} \;\; W_{ijk}^{-++} \\
f_{ij}^{+-} + f_{jk}^{+-} & = & f_{ik}^{++} \hspace{1pc} \text{on} \;\; W_{ijk}^{+-+} \\
f_{ij}^{++} + f_{jk}^{+-} & = & f_{ik}^{+-} \hspace{1pc} \text{on} \;\; W_{ijk}^{++-}
\end{eqnarray*}
This yields the curiously looking identities
\begin{eqnarray*}
f_{ij}^{++} + f_{jk}^{++}  & = & f_{ij}^{+-} + f_{jk}^{+-} \\
f_{ij}^{+-} + f_{jk}^{++} & = & f_{ij}^{++} + f_{jk}^{+-}.
\end{eqnarray*}
Subtracting these above equations gives the condition that $2\left (f_{ij}^{++} - f_{ij}^{+-}\right ) = 0$, but the question is whether the original work I did above enough to use this as justification.  I have a feeling at this point that this is the desired result, but I'm honestly lost by my own computation.  Is this proof that elements of $H^1(\mathcal{U}, \mathcal{A})$ just the elements of $A$ generated by $f_{ij}^{++} - f_{ij}^{+-}$ and that this computation shows that these elements must be of order 2?
 A: I believe I can answer the main question I posed at the end: that is, whether it's sufficient to say $H^1\left (\mathbb{RP}^n, \mathcal{A}\right ) = A/2A$ for $n\geq 2$.  If correct, the work above shows that the only cocycles that can exist on $\mathbb{RP}^n$ are those that when restricted to this standard covering of coordinate charts satisfy that $f_{ij}^{++} - f_{ij}^{+-}$ is of order 2: as shown in edit #2 above.  The work I did in edit #1 above is still valid, but I used different notations.  One could instead say that $\left \{\left (f_{ij}^{++}, f_{ij}^{+-}\right )\right \}$ and $\left \{\left (g_{ij}^{++}, g_{ij}^{+-}\right )\right \}$ are cohomologous if
$$
f_{ij}^{++} - f_{ij}^{+-} \;\; =\;\; g_{ij}^{++} - g_{ij}^{+-}
$$
for all $i,j$.  Notice however these are elements of order 2 in $A$, and so it is all of the differing elements of order 2 in $A$ that generate this cohomology class.  Thus
$$
H^1\left (\mathbb{RP}^n, \mathcal{A}\right ) \;\; =\;\; H^1(\mathcal{U}, \mathcal{A}) \;\; =\;\; A/2A.
$$
